Consider two hydrogen atoms `H_(A)andH_(B)` in ground state. Assume that hydrogen atom `H_(A)` is at rest and hydrogen atom `H_(B)` is moving with a speed and make head-on collide on the stationary hydrogen atom `H_(A)`. After the strike, both of them move together. What is minimum value of the kinetic energy of the moving hydrogen atom `H_(B)`, such that any one of the hydrogen atoms reaches one of the excitation state.

Explanation

`H_(A)-"rest".u_(2)=0`
`H_(B)` - moving with a speed `u_(1)`
After the strike, velocity of `H_(A)&H_(B)` is same
i.e. `v_(1)=v_(2)=v`
Mass of `H_(A)` = mass of `H_(B)` = m
Total Kinetic energy before collission
Kinetic energy = `1/2m u_(1)^(2)+1/2m u_(2)^(2)=1/2m u_(1)^(2)" "(u_(2)=0)`
Total kinetic energy after collission
Kinetic energy = `1/2(m+m)v^(2)`
= `1/2*2mV^(2)=mv^(2)`
Less in kineetic energy = `K.E_(f)-K.E_(f)`
`mv^(2)-1/2m u_(1)^(2)" "...(1)`
`DeltaE=1/2(m^(2)/(2m))u_(1)^(2)" "["substitute "v=(m u_(1))/(2m)]` in equation (1)
`DeltaE=1/4m u_(1)^(2)`
`2DeltaE=1/2m u_(1)^(2)`
The minimum energy that can be absorbed by the `H_(2)` alom in ground state to go to an excited state is 10.2 eV.
Thus minimum kinetic energy the moving `H_(2)` atom `H_(B)`.
`1/2m u_(min)^(2)=2xx10.2=20.4eV`.